ΔABC is a right angled triangle, ∠C - 90 degree, Ab = 25 cm and BC = 20 cm. What is the value of sec A ?

$\frac{5}{3}$ $\frac{4}{5}$ $\frac{4}{3}$ $\frac{5}{4}$

$\frac{5}{3}$

We have, AB = 25 cm, BC = 20 cm ∠C = 90°, Using Pythagoras Theorem AC = \(\sqrt {(AB^2 – BC^2)}\) AC = \(\sqrt {(25^2 – 20^2)}\) = 15 cm = sec A = \(\frac{ AC }{ BC }\) = \(\frac{ 25 }{ 15 }\) = \(\frac{ 5 }{ 3 }\)